Chapter 14 · Surface integrals and Stokes' theorem

The hook

A solar panel, and the sign integration grew

Point a solar panel straight at the sun and it harvests the most power it can. Tilt it and the yield falls off. Turn it edge-on — its face parallel to the incoming light — and it collects nothing, even though the sun is blazing at full strength. The power a panel collects is the field of sunlight dotted with the panel's normal direction: $\mathbf F\cdot\hat{\mathbf n}$ , and that dot product falls as $\cos\theta$ as you tilt.

Same field, same panel — four fluxes. Integration just grew a sign that depends on a choice you make (which way the normal points), and a value that vanishes edge-on though the field never weakened. School integration never did either of those.

The same dot product is how 3-D graphics decides how bright a surface looks — Lambert shading is $\mathbf F\cdot\hat{\mathbf n}$ between a light direction and a surface normal, the exact computation in the renderers behind every game and animated film. This chapter is the calculus of that idea: integrating a quantity over a curved surface, and the orientation choice that gives the answer its sign.

Why this matters

A surface integral totals something spread over a curved sheet — its area, its mass, or the flux of a field through it (how much flows across). And just as Chapter 13's Green's theorem traded a boundary loop for the region inside, Stokes' theorem trades a loop in 3-D space for a flux through any surface that loop bounds. One honest warning up front: nobody pictures a curved surface whole. We parametrise it down to a flat $(u,v)$ rectangle and integrate there — the 3-D version of Chapter 10's shadow-and-slice.

The ritual, lifted to a surface

Parametrising a surface, and the element $dS$

The five-step spine you have run since Chapter 10 does not change — only two of its steps swap out. A solid was described by sketch → shadow → slice. A surface is described by parametrise → build the normal → orient:

	1	2	3	4	5
Solids (Ch 10–13)	sketch	shadow	slice	limits	integrate
Surfaces (Ch 14)	parametrise	r_u × r_v	orient	(u,v) limits	integrate

Three steps keep their role (draw/describe, read the limits, integrate). The two solid-only steps — shadow and slice — swap for the two surface-only steps: build the normal with a cross product, and orient it. "Orient" is a named guardrail, because orientation is exactly where marks leak.

A surface is a map $\mathbf r(u,v) = \big(x(u,v), y(u,v), z(u,v)\big)$ from a flat domain $D$ . The two tangent vectors $\mathbf r_u$ and $\mathbf r_v$ span the little patch, and their cross product does triple duty:

\mathbf r_u \times \mathbf r_v

— one vector, three jobs

It is perpendicular to the surface — so it is the normal direction.
Its length is the area of the little parallelogram the patch spans — so $|\mathbf r_u \times \mathbf r_v|$ is exactly $dS$ per unit $du\,dv$ . (It is the Jacobian of the surface map — the same area-scaling machine as Chapters 11 and 12, now mapping a flat rectangle onto a curved sheet.)
Its direction is the normal that carries the orientation — the sign of the flux.

The surface element, and the graph special case

dS = |\mathbf r_u \times \mathbf r_v|\,du\,dv, \qquad \text{for } z=f(x,y):\ \ dS = \sqrt{1 + f_x^2 + f_y^2}\,dA, \ \ \mathbf r_x\times\mathbf r_y = (-f_x, -f_y, 1).

The graph case $z = f(x,y)$ is the most common exam surface, and there the $(u,v)$ domain is the $xy$ -shadow of the surface — the one place the old word "shadow" still applies literally.

Worked example · WE1 · the machine recovers

4\pi a^2

Find the surface area of a sphere of radius $a$ using $\mathbf r(\phi,\theta) = (a\sin\phi\cos\theta,\ a\sin\phi\sin\theta,\ a\cos\phi)$ . The cross product of the tangent vectors works out to

|\mathbf r_\phi \times \mathbf r_\theta| = a^2\sin\phi, \qquad \text{Area} = \int_0^{2\pi}\!\!\int_0^{\pi} a^2\sin\phi\,d\phi\,d\theta = a^2\cdot 2\cdot 2\pi = 4\pi a^2.

The element $a^2\sin\phi$ is the spherical surface Jacobian — the same $\sin\phi$ that ran through Chapter 12's volume element, now without the $\rho^2$ . The machine recovers the area you already trust; a hemisphere ( $\phi: 0 \to \tfrac{\pi}{2}$ ) gives $2\pi a^2$ .

Totalling a quantity over a sheet

Scalar surface integrals

With $dS$ in hand, a scalar surface integral is just "sum $f$ over the sheet":

Scalar surface integral

\iint_S f\,dS = \iint_D f\big(\mathbf r(u,v)\big)\,|\mathbf r_u \times \mathbf r_v|\,du\,dv.

Take $f = 1$ and you get area; $f =$ density and you get the mass of a curved sheet (a dome, a shell, a sail); divide by area and you get an average value. There is no orientation here — a scalar integral uses the length $|\mathbf r_u\times\mathbf r_v|$ , which is always positive.

Worked example · WE2 · mass-style integral over a dome

Evaluate $\displaystyle\iint_S z\,dS$ over the upper hemisphere $x^2 + y^2 + z^2 = a^2$ , $z \ge 0$ . On the sphere $z = a\cos\phi$ and $dS = a^2\sin\phi\,d\phi\,d\theta$ , with $\phi: 0 \to \tfrac{\pi}{2}$ :

\iint_S z\,dS = \int_0^{2\pi}\!\!\int_0^{\pi/2}(a\cos\phi)(a^2\sin\phi)\,d\phi\,d\theta = 2\pi a^3\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi = 2\pi a^3\cdot\tfrac12 = \pi a^3.

The same $\int\cos\phi\sin\phi = \tfrac12$ you met computing the hemisphere's centroid in Chapter 12 — the machinery is genuinely shared across the unit.

End of sitting 1 — you can parametrise a surface, build $dS$ from the cross product (and the graph shortcut $\sqrt{1+f_x^2+f_y^2}$ ), and integrate a scalar over it. After the break: flux and the orientation sign that trips everyone, then Stokes' theorem — Green's theorem lifted off the plane — with two playgrounds to drill both.

The mark-loser: orientation

Flux, and the sign that follows your choice

Flux measures how much of a field crosses a surface. It is the field dotted with the normal, summed over the sheet — and because the normal keeps its direction this time, the vector cross product is used, sign and all:

Flux of

\mathbf F

through

S

\iint_S \mathbf F\cdot\hat{\mathbf n}\,dS = \iint_D \mathbf F\cdot(\mathbf r_u \times \mathbf r_v)\,du\,dv.

A surface has two normals — the patch can face either way — and choosing one is choosing the sign. "Outward" (for a closed surface) and "upward" (for a graph) are the usual conventions, but the problem statement decides. The flux sign follows the normal you pick — this is the chapter's headline error.

Worked example · WE3 · the same flux, both orientations

Find the flux of $\mathbf F = (x, y, z)$ through the paraboloid cap $z = 1 - x^2 - y^2$ , $z \ge 0$ . As a graph, the upward normal vector is $\mathbf N = (-f_x, -f_y, 1) = (2x, 2y, 1)$ , so on the surface $\mathbf F\cdot\mathbf N = 2x^2 + 2y^2 + z = 2x^2 + 2y^2 + (1 - x^2 - y^2) = x^2 + y^2 + 1$ . Polar over the unit-disk shadow:

\text{(upward)}\quad \int_0^{2\pi}\!\!\int_0^{1}(r^2 + 1)\,r\,dr\,d\theta = 2\pi\Big(\tfrac14 + \tfrac12\Big) = \frac{3\pi}{2}, \qquad \text{(downward)}\quad -\frac{3\pi}{2}.

Identical surface, identical field — the only difference is which way $\hat{\mathbf n}$ points, and that flips the sign. Quote a flux without stating the orientation and the number is only half an answer.

Playground 1 — Flux Tilt

A flat panel of area $A$ sits in a uniform field of strength $F_0$ . Drag the panel's tilt $\theta$ and toggle which way its normal points, then — before you reveal — predict the sign of the flux. Watch it vanish edge-on, and flip when you flip the normal.

Flux Tilt · Φ = sign · F₀ · A · cos θ

Panel & field:

Tilt θ: θ = 0°

Normal:

Predict the flux:

field meets panel

Edge-on (θ = 90°) the flux is 0 in every preset — the field grazes the panel and nothing crosses, though the field is at full strength. The minus normal flips the sign; θ = 0° gives the maximum ±F₀·A.

Green's theorem, lifted off the plane

Stokes' theorem

In 3-D the curl becomes a vector — it measures the axis and rate of a field's local spinning:

3-D curl

\operatorname{curl}\mathbf F = \nabla\times\mathbf F = (R_y - Q_z,\ \ P_z - R_x,\ \ Q_x - P_y), \qquad \mathbf F = (P, Q, R).

Chapter 13's flat scalar curl $Q_x - P_y$ was just the $\mathbf k$ -component of this — the part that spins in the plane. Stokes' theorem then says:

Stokes' theorem

\oint_C \mathbf F\cdot d\mathbf r = \iint_S (\operatorname{curl}\mathbf F)\cdot\hat{\mathbf n}\,dS.

In words: a field's circulation around a closed space curve $C$ equals the flux of its curl through any surface that the curve bounds, with $C$ oriented by the right-hand rule about $\hat{\mathbf n}$ . Flatten everything to the plane ( $\hat{\mathbf n} = \mathbf k$ ) and this is Green's theorem from Chapter 13. Stokes is Green lifted off the plane.

Worked example · WE4 · one boundary, three surfaces, one answer

Take $\mathbf F = (-y, x, 0)$ , so $\operatorname{curl}\mathbf F = (0, 0, 2)$ , and let $C$ be the unit circle in the $xy$ -plane, counterclockwise.

(a) Direct circulation

$\displaystyle\oint_C (-y\,dx + x\,dy) = 2\pi$ (the Chapter 13 area form).

(b) Stokes via the flat disk

$\hat{\mathbf n} = \mathbf k$ : $\displaystyle\iint_S 2\,dS = 2\cdot\pi = 2\pi.$ ✓

Same rim, outward normal: $\displaystyle\iint_S (0,0,2)\cdot\hat{\mathbf n}\,dS = \int_0^{2\pi}\!\!\int_0^{\pi/2} 2\cos\phi\sin\phi\,d\phi\,d\theta = 2\pi.$ ✓

Same value — the bulging dome and the flat disk agree.

The curl-flux did not care which surface you stretched across the loop — only the boundary. It is the same lesson as Chapter 13's "the integral sees only what the contour encloses," now one dimension up.

Playground 2 — Stokes Verifier

The field is fixed at $\mathbf F = (-y, x, 0)$ . Pick a surface stretched across a rim, predict its curl-flux, and reveal it next to the boundary circulation $\oint_C\mathbf F\cdot d\mathbf r$ . Swap to a different surface on the same rim — the value does not budge. Only changing the rim changes it.

Stokes Verifier · curl-flux = 2 · area(rim) = ∮ F·dr

Surface (unit rim → 2π):

Surface (radius-2 rim → 8π):

Predict the curl-flux:

surface & rim

All four unit-rim surfaces return 2π; the radius-2 surfaces return 8π. The value tracks the rim, never the surface — that is surface independence, the heart of Stokes.

Retrieval gauntlet

Test yourself before moving on

Commit before reading the feedback. Two of these turn on the orientation sign — the exact place marks leak — on purpose.

Quiz 1 · the

dS

machinery

Using a parametrisation, the surface area of a sphere of radius $2$ is:

Quiz 2 · orientation sign

The flux of $\mathbf F = (x, y, z)$ through the paraboloid cap $z = 1 - x^2 - y^2$ , $z \ge 0$ , oriented downward, is:

Quiz 3 · Stokes' value

For $\mathbf F = (-y, x, 0)$ and $C$ the unit circle CCW, Stokes' theorem gives $\displaystyle\oint_C \mathbf F\cdot d\mathbf r =$

Quiz 4 · which tool? — decide, don't compute

$C$ is a messy closed loop in the $xy$ -plane (awkward to parametrise) bounding a region of known area. To find $\oint_C \mathbf F\cdot d\mathbf r$ for $\mathbf F = (-y, x, 0)$ , the efficient route is:

Quiz 5 · orientation in the boundary

The same $\mathbf F = (-y, x, 0)$ , but with $C$ the unit circle taken clockwise (seen from $+z$ ): $\displaystyle\oint_C \mathbf F\cdot d\mathbf r =$

Exam translator — what a surface/Stokes question actually asks

Exam shapes

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, this material asks four things:

"Set up / evaluate $\iint_S f\,dS$ " → parametrise, build $|\mathbf r_u\times\mathbf r_v|$ (or $\sqrt{1+f_x^2+f_y^2}$ for a graph), integrate over $D$ (WE1, WE2).
"Find the flux of $\mathbf F$ through $S$ (oriented …)" → use the vector $\mathbf r_u\times\mathbf r_v$ ; the sign follows the stated orientation (WE3). This is where marks leak.
"Verify Stokes' theorem for …" → compute both the loop integral and the curl-flux and show they match (WE4).
"Using Stokes', evaluate …" → turn a hard loop integral into an easy curl-flux, or vice versa (Q4).

Worked solutions are complete; nothing points off this page. Do them as listed — interleaved on purpose. PR7 opens with a faded bridge: part of the work is set out, the key step left for you.

PR1 · surface area of a cone (graph)

Shape · graph z = f(x,y) ⇒ dS = √(1+f_x²+f_y²) dA

PR1. Find the area of the cone $z = \sqrt{x^2 + y^2}$ below $z = 1$ .

For $f = \sqrt{x^2+y^2}$ , $f_x^2 + f_y^2 = \dfrac{x^2+y^2}{x^2+y^2} = 1$ , so the $dS$ factor is $\sqrt{1+1} = \sqrt{2}$ — constant. The shadow is the unit disk:

\text{Area} = \iint_{D}\sqrt{2}\,dA = \sqrt{2}\cdot\pi = \pi\sqrt{2}.

PR2 · scalar integral over that cone

Shape · ∬ f dS with f = z, the same √2 factor

PR2. Evaluate $\displaystyle\iint_S z\,dS$ over that cone, $0 \le z \le 1$ .

On the cone $z = r$ and $dS = \sqrt{2}\,r\,dr\,d\theta$ :

\iint_S z\,dS = \int_0^{2\pi}\!\!\int_0^{1} r\cdot\sqrt{2}\,r\,dr\,d\theta = 2\pi\sqrt{2}\cdot\tfrac13 = \frac{2\sqrt{2}\,\pi}{3}.

PR3 · flux through a cylinder side (both orientations)

Shape · parametric surface; outward normal, then flip

PR3. Find the flux of $\mathbf F = (x, y, z)$ through the side of the cylinder $x^2 + y^2 = 1$ , $0 \le z \le 2$ , oriented outward.

Parametrise $(\cos\theta, \sin\theta, z)$ ; the outward normal is $\hat{\mathbf n} = (\cos\theta, \sin\theta, 0)$ , so on the side $\mathbf F\cdot\hat{\mathbf n} = \cos^2\theta + \sin^2\theta = 1$ , and $dS = d\theta\,dz$ :

\iint_S \mathbf F\cdot\hat{\mathbf n}\,dS = \int_0^{2}\!\!\int_0^{2\pi} 1\,d\theta\,dz = 4\pi.

Inward orientation flips it to $-4\pi$ . (The flat top and bottom caps are not part of "the side.")

PR4 · Stokes, where the sign rides the field

Shape · curl from the field, not the orientation this time

PR4. For $\mathbf F = (y, -x, 0)$ and $C$ the unit circle CCW, find $\displaystyle\oint_C \mathbf F\cdot d\mathbf r$ .

$\operatorname{curl}\mathbf F = (0, 0, Q_x - P_y) = (0, 0, -1 - 1) = (0, 0, -2)$ . The curl-flux through the disk is $-2\cdot\text{area} = -2\pi$ . Here the sign comes from the field (the curl points down), not from the orientation — contrast Q5, where the field was $(-y, x, 0)$ and the sign came from walking $C$ backwards.

PR5 · flux through a flat disk (orientation)

Shape · constant field, flat disk, stated orientation

PR5. Find the flux of $\mathbf F = (0, 0, 3)$ through the disk $x^2 + y^2 \le 4$ in the plane $z = 1$ , oriented downward.

The disk has area $\pi\cdot 2^2 = 4\pi$ . With the upward normal, $\mathbf F\cdot\hat{\mathbf n} = 3$ , giving $3\cdot 4\pi = 12\pi$ . Oriented downward, the sign flips:

\iint_S \mathbf F\cdot\hat{\mathbf n}\,dS = -12\pi.

PR6 · decision-only · which tool, and the value

Shape · no parametrisation — Stokes/Green by inspection

PR6. $C$ bounds a region of area $7$ . For $\mathbf F = (-y, x, 0)$ , name the efficient tool and give $\oint_C \mathbf F\cdot d\mathbf r$ .

Stokes / Green. $\operatorname{curl}\mathbf F = (0,0,2)$ is constant, so the circulation $= 2\cdot\text{area} = 2\cdot 7 = 14$ — no parametrisation needed.

PR7 · faded bridge · surface independence, confirmed

Shape · same rim, two surfaces ⇒ same curl-flux

PR7. For $\mathbf F = (-y, x, 0)$ , confirm surface independence by computing $\displaystyle\iint_S (\operatorname{curl}\mathbf F)\cdot\hat{\mathbf n}\,dS$ over both the flat disk and the paraboloid cap $z = 1 - x^2 - y^2$ (both with the unit-circle rim).

Bridge: $\operatorname{curl}\mathbf F = (0,0,2)$ . For the disk $\hat{\mathbf n} = \mathbf k$ ; for the paraboloid the upward $\mathbf N = (2x, 2y, 1)$ . Predict the next step: what is $(0,0,2)\cdot\mathbf N$ , and why does the integral come out the same?

Disk: $(0,0,2)\cdot\mathbf k = 2$ , so $\iint 2\,dA = 2\pi$ . Paraboloid: $(0,0,2)\cdot(2x, 2y, 1) = 2$ as well — the curl's $x,y$ components are zero, so only the $\mathbf k$ -part survives, and the integrand is again $2$ over the unit disk:

\iint_S (\operatorname{curl}\mathbf F)\cdot\hat{\mathbf n}\,dS = \iint_D 2\,dA = 2\pi \quad\text{both ways.}

The curl had no sideways component, so the surface's bulge contributed nothing — only the rim mattered. That is surface independence in one line.

Pre-flight checklist — run it before you integrate

\textbf{1.}\ \text{Surface integral: is } dS = |\mathbf r_u\times\mathbf r_v|\,du\,dv\ (\text{not plain } dA)?\ \text{Graph: the } \sqrt{1+f_x^2+f_y^2}.

\textbf{2.}\ \text{Flux: is the normal the one the problem names (outward / upward / downward)? The sign follows it.}

\textbf{3.}\ \text{Scalar uses } |\mathbf r_u\times\mathbf r_v| \text{ (magnitude); flux keeps the vector } \mathbf r_u\times\mathbf r_v \text{ (sign).}

\textbf{4.}\ \text{Stokes: does } C\text{'s direction match } \hat{\mathbf n} \text{ by the right-hand rule? Mismatch = wrong sign.}

Three things to carry out of this chapter

Parametrise, then the cross product does everything. $|\mathbf r_u\times\mathbf r_v|$ is the area element $dS$ (the surface Jacobian); its direction is the normal. Graphs get the shortcut $dS = \sqrt{1+f_x^2+f_y^2}\,dA$ .
Flux carries a sign, and the sign is your choice of normal. Scalar integrals use the magnitude; flux keeps the vector. State the orientation or the answer is half done.
Stokes is Green lifted off the plane: $\oint_C \mathbf F\cdot d\mathbf r = \iint_S (\operatorname{curl}\mathbf F)\cdot\hat{\mathbf n}\,dS$ , and the curl-flux depends only on the boundary $C$ — any surface with the same rim gives the same value.

The anchor, one more time

Back to the solar panel: the power it harvests is $\mathbf F\cdot\hat{\mathbf n}$ summed over its face — a flux integral — and edge-on it collects nothing, the same $\cos\theta$ that opened the chapter. Orientation is not a technicality; it is the difference between a panel that works and one that doesn't. Re-do PR3 and PR7 from a blank page tomorrow — not re-read, re-do. Keeping the normal's direction straight is the muscle this chapter builds.

What's next. Chapter 15 closes the course with the divergence theorem — flux out through a closed surface equals the divergence summed over the solid inside — and then steps back to see Green's, Stokes', and Gauss's theorems as three faces of one idea: the boundary remembers the inside.

Same material, another voice

If a different explanation would help, this maps onto the chapter — free, from MIT OpenCourseWare:

Read / watch: MIT 18.02SC — Part C: Line Integrals and Stokes' Theorem — frames Stokes as the extension of Green's, with the curl-flux over any surface sharing the boundary (exactly WE4 and the Stokes Verifier).

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.

A solar panel, and the sign integration grew

Parametrising a surface, and the element dSdSdS

Scalar surface integrals

Flux, and the sign that follows your choice

Playground 1 — Flux Tilt

Stokes' theorem

Playground 2 — Stokes Verifier

Test yourself before moving on

Exam translator — what a surface/Stokes question actually asks

PR1 · surface area of a cone (graph)

PR2 · scalar integral over that cone

PR3 · flux through a cylinder side (both orientations)

PR4 · Stokes, where the sign rides the field

PR5 · flux through a flat disk (orientation)

PR6 · decision-only · which tool, and the value

PR7 · faded bridge · surface independence, confirmed

Three things to carry out of this chapter

Same material, another voice

Parametrising a surface, and the element $dS$